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Veseljchak [2.6K]
3 years ago
13

A book is pushed off the edge of a table. At what point during the fall will the potential and kinetic energy be equal?

Chemistry
1 answer:
vfiekz [6]3 years ago
5 0

Answer:

KE = PE at half the table Height:

Explanation

AT ANY POINT IN THE BOOK'S FALL,

TOTAL E = PE +KE

THE TOTAL E IS CONSTANT

Before the book is pushed off, the total energy is potential

TOT E=PE =MGH

BEFORE THE BOOK HITS THE GROUND, THE TOTAL E IS KINETIC

TOT=KE = MVXV/2

WHEN KE = PE

KE+PE =<u> MGH (STARTING ENERGY SINCE E IS CONSERVED)</u>

<u>OR PE+ PE = MGH</u>

<u>OR MGH' + MGH' =MGH</u>

<u>OR 2H' =</u>H

H' (NEW HEIGHT) =H/2

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Calculate the maximum numbers of moles and grams of H₂S that can form when 158 g of aluminum sulfide reacts with 131 g of water:
Phantasy [73]

What is Chemical Reaction?

A chemical reaction is the chemical transformation of one set of chemical components into another.

Main Content

Mass of aluminium sulfide is 158g

Mass of water is 131g

The chemical reaction: Al_{2}S_{3} +H_{2}O  _\to  Al(OH)_{3} + H_{2}S

First, balance the chemical equation

Al_{2}S_{3} + 6H_{2}O  \to 2Al(OH)_{3} + 3H_{2}S

Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:

n_{Al_{2}S_{3}  } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3}  } = \frac{158g}{150.16g/mol}   \\n_{Al_{2}S_{3} }=1.05 mol

From the chemical reaction , the ratio of molar is 3mol H_{2}S/1 mol Al_{2}S_{3}. So, the moles of hydrogen sulfide are:

n_{H_{2} O} =\frac{131g}{18.02g/mol}

       = 7.26mol

From the chemical reaction, the molar ratio is 3 mol H_{2}S/6 mol H_{2}O. So, the moles of hydrogen sulfide are:

Moles of H_{2}S formed = 7.26 mol H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }

Th liming reactant isAl_{2}S_{3} beacuse the mass of Al_{2}S_{3} forms less product than water. Therefore, the maximum number of moles of H_{2}S is 3.15 mol.  We know that molar mass of H_{2}S is 34.10g/mol. So, the maximum mass of H_{2}S formed is,

m_{H_{2}S } = n_{H_{2}S } \times Molar mass of H_{2}S

         = 3.15 mol \times 34.10g/mol

         = 107.4g

Now, multiplying the number of moles of Al_{2}S_{3} by the molar ratio between Al_2S_3 and H_2O which is 6mol H_2O/1mol Al_2S_3 we get the number of moles of H_2O reacted.

Moles of H_2O reacted = 1.05 mol Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}

                                     = 6.31 mol H_2O

The mass of H_2O is,

m_{H_{2} O} = 6.31 mol \times 18.02g/ mol

          = 114g

On subtracting, the mass of H_2O reacted from the given mass of H_2O is,

m_{H_2O} = (131-114)g

         = 17g

Hence, the excess remaining reactant is 17g

To learn more about Chemical Reaction

brainly.com/question/11231920

#SPJ4

 

8 0
2 years ago
Please help with both
kupik [55]

Answer:

D =Average atomic mass  = 10.801 amu.

5) True

Explanation:

Abundance of B¹⁰= 19.9%

Abundance of B¹¹ = 80.1%

Atomic mass of B¹⁰ = 10 amu

Atomic mass of B¹¹ = 11 amu

Average atomic mass = ?

Solution:

Average atomic mass  = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

Average atomic mass  = (10×19.9)+(11×80.1) /100

Average atomic mass =  199 + 881.1 / 100

Average atomic mass  = 1080.1 / 100

Average atomic mass  = 10.801 amu.

2)A chemical reaction is one in which a new elements is created

True

False

Answer:

In chemical reaction new substances are created.

For example:

Photosynthesis:

It is the process in which in the presence of sun light and chlorophyll by using carbon dioxide and water plants produce the oxygen and glucose.

Carbon dioxide + water + energy →   glucose + oxygen

water is supplied through the roots, carbon dioxide collected through stomata and sun light is capture  by chloroplast.

Chemical equation:

6H₂O + 6CO₂ + energy  →   C₆H₁₂O₆ + 6O₂

it is known from balanced chemical equation that 6 moles of carbon dioxide react with the six moles of water and created one mole of glucose and six mole of oxygen.

8 0
3 years ago
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